Question

Let $f\left(x\right)=({\sin }^{-1}x{)}^2-({\cos }^{-1}x{)}^2.$ If range of $f$ equals $[\frac{a{\pi }^2}{4},\frac{b{\pi }^2}{4}]$ where $a,b\in Z,$ then the value of $b-a$ is

Answer 1

Given,
$f\left(x\right)=({\sin }^{-1}x{)}^2-({\cos }^{-1}x{)}^2=({\sin }^{-1}x+{\cos }^{-1}x)({\sin }^{-1}x-{\cos }^{-1}x)=\frac{\pi }{2}({\sin }^{-1}x-(\frac{\pi }{2}-{\sin }^{-1}x))\therefore f\left(x\right)=\frac{\pi }{2}(2{\sin }^{-1}x-\frac{\pi }{2}),\forall x\in [-1,1]$
As $f$ is increasing function,
so, $f_{max}=f\left(1\right)=\frac{\pi }{2}(2\left(\frac{\pi }{2}\right)-\frac{\pi }{2})=\frac{{\pi }^2}{4}$
and $f_{min}=f(-1)=\frac{\pi }{2}(2(-\frac{\pi }{2})-\frac{\pi }{2})=\frac{-3{\pi }^2}{4}$
Hence, range of $f$ is$[\frac{-3{\pi }^2}{4},\frac{{\pi }^2}{4}]$
$\therefore a=-3$ and $b=1$
$\therefore b-a=1-(-3)=4$